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# Further results on common zeros of the solutions of two differential equations

*Journal of Inequalities and Applications*
**volume 2012**, Article number: 222 (2012)

## Abstract

### Purpose

Two problems are discussed in this paper. In the first problem, we consider one homogeneous and one non-homogeneous differential equations and study when the solutions of these differential equations can have (nearly) the same zeros. In the second problem, we consider two linear second-order differential equations and investigate when the solutions of these differential equations can take the value 0 and a non-zero value at (nearly) the same points.

### Method

We apply the Nevanlinna theory and properties of entire solutions of linear differential equations.

### Conclusion

In the first problem, the results determine all pairs of such equations having solutions with the same zeros or nearly the same zeros. Regarding the second problem, the results also show all pairs of such equations having solutions taking the value 0 and a non-zero value at (nearly) the same points.

## 1 Introduction

There has been much research [1–8] on zeros of solutions of linear differential equations with entire coefficients. The principal paper [9] that was published in 1982 by Bank and Laine has stimulated many studies on this kind of problems. The reader is referred to [10–12] for background on some applications of the Nevanlinna theory. We use the standard notation of the Nevanlinna theory from [13].

In 1955, Wittich [12] proved the following theorem.

**Theorem 1.1** *If* *f* *is a non*-*trivial solution of* ${w}^{\u2033}+Aw=0$, *i*.*e*., $f\not\equiv 0$ *and* $A\not\equiv 0$ *is entire*, *then we have*:

(i) $T(r,A)=S(r,f)$.

(ii) *If* *f* *has finite order*, *then* *A* *is a polynomial*.

(iii) *If* *a* *is a non*-*zero complex number*, *then* *f* *takes the value* *a* *infinitely often*, *and in fact*, *outside a set of finite measure*,

The following facts follow from the asymptotic representation for solutions of the equation

*Let* *P* *be a polynomial of degree* *n*, *and let* *w* *be a non*-*trivial solution of the equation* (1). *Then*, *w* *has order of growth equal to* $\frac{n+2}{2}$. *Moreover*, *if* *w* *is a solution of* (1) *which has infinitely many zeros*, *then we have*

Our previous paper [14] studied homogeneous linear differential equations having solutions with nearly the same zeros and proved several results, including the following.

**Theorem 1.3** [14]

*Let* $P\not\equiv 0$ *be a polynomial of degree* *n*. *Let* $w\not\equiv 0$ *be a solution of* (1). *Assume that* *w* *has infinitely many zeros*. *Suppose that we have a solution* $v\not\equiv 0$ *of the differential equation*

*such that* *A* *is an entire function and* $N(r)$ *counts zeros of* *v* *which are not zeros of* *w* *and zeros of* *w* *which are not zeros of* *v*. *Assume that*

*Then* $\frac{v}{w}$ *is a constant and* $A=P$.

The paper [14] includes further results for homogeneous linear differential equations, and the corresponding problem where *P* is a transcendental entire function of finite order is studied in [15].

Recently, the same problem, but with non-homogeneous first-order differential equations, has been studied in [16], including the following result.

**Theorem 1.4** [16]

*Assume that* ${v}^{\prime}=Av+B$ *and* ${w}^{\prime}=Cw+D$, *where* *A*, *B*, *C* *and* *D* *are entire functions of order less than* 1 *and* *v*, *w* *are transcendental functions*. *Assume that* $v=Lw$, *where* *L* *has finitely many zeros and poles*, *and*

*Then the following conclusions hold*.

(I) *If* *L* *is a rational function*, *then* $A\equiv C$, *L* *is a constant and* $B=LD$.

(II) *If* *L* *is a transcendental function*, *then one of the following cases holds*: *If*, *in addition*, *L* *has finite order in* case (ii), *then* *A*, *B*, *C*, *D* *are polynomials and so is* ${A}_{1}$.

(i) $B\equiv D\equiv 0$ *and* *v*, *w* *have no zeros*.

(ii) $A=-C$ *and* $B/A$, $D/C$ *are non*-*zero constants*, *and*

*where* ${c}_{j}\in \mathbb{C}$, ${A}_{1}^{\prime}=A$ *and* $L=(\mathit{\text{constant}}){e}^{{A}_{1}}$.

In this paper, our first result (Theorem 2.1) looks at the same problem but with one homogeneous and one non-homogeneous differential equations. In particular, we consider the first equation to be homogeneous of the second-order with a polynomial coefficient and the second equation to be non-homogeneous of the first-order with entire coefficients.

A further result (Theorem 2.2) studies the case where the solutions of two second-order homogeneous differential equations can take the value 0 and a non-zero value at (nearly) the same points.

## 2 Our results

Our first result is the following theorem.

**Theorem 2.1** *Suppose that* $P\not\equiv 0$ *is a polynomial of degree* *n*, *and* *w* *solves* (1), *and* $w\not\equiv 0$ *has infinitely many zeros*. *Suppose that* $v\not\equiv 0$ *solves*

*where* *A*, *B* *are entire and* $AB\not\equiv 0$, *and*

*Suppose that* $L=\frac{w}{v}$ *has finitely many zeros and poles* (*i*.*e*., *w* *and* *v* *have the same zeros with finitely many exceptions*).

*Then*
*A*
*is a polynomial and there exists a polynomial*
*Q*
*such that*

*and*

*where* ${A}_{1}^{\prime}=A$ *and* ${c}_{1}$, ${c}_{2}$ *are constants*.

**Example 2.1** Take *Q* to be a polynomial. Let

Then

and

So, we have $P=-({Q}^{\u2033}+{Q}^{\mathrm{\prime}2})$.

Now, let ${A}_{1}$ be another polynomial, and let

Note that *v* has the same zeros as *w*. Now, we have

where $A={A}_{1}^{\prime}$.

We choose ${A}_{1}$ so that

For example, let ${A}_{1}=2Q$.

We now state our second result.

**Theorem 2.2** *Suppose that* $P\not\equiv 0$ *is a polynomial of degree* *n*, *and* *A* *is an entire function*, *and suppose that* *w* *solves* (1) *and* *v* *solves* (3), *and* $vw\not\equiv 0$. *Let* $v-1$ *and* *w* *have*, *with finitely many exceptions*, *the same zeros and the same multiplicities*. *Then one of the following holds*.

(A) *w* *has finitely many zeros and* *v* *is a polynomial and* $A=0$.

(B) *w* *has infinitely many zeros and* *P*, *A* *are non*-*zero constants and* $\frac{v-1}{w}$ *is non*-*constant and*

*where* *σ*, ${\lambda}_{1}$, ${\lambda}_{2}$, ${\lambda}_{3}$ *are non*-*zero constants*.

**Example 2.2** If $w={e}^{z}-{e}^{-z}$ and $v={e}^{2z}$, then

Hence, $v-1$ has the same zeros as *w*. Here $P=-1$ and $A=-4$.

**Example 2.3** We give an example to show that the zeros of $v-1$ and *w* must necessarily have the same multiplicities. To show this, let

Then $w=0\iff \frac{z}{2}=k\pi $, where $k\in \mathbb{Z}$.

Also $v=1\iff z=k2\pi $, where $k\in \mathbb{Z}$.

Therefore, *w* and $v-1$ have the same zeros but the zeros are simple for *w*, double for $v-1$. Here, $P=\frac{1}{4}$ and $A=1$.

## 3 Proof of Theorem 2.1

*Proof* We have

So,

but

We also have, using (1), (5), (8), (9) and (10),

Let

Then

We divide (11) by *L*, and by using (12) and (13), we get

The next step is to estimate the growth of *M*.

We know that $\rho (w)=\frac{n+2}{2}$ from [9]. Therefore,

**Claim 3.1** *We claim that* $T(r,M)=o({r}^{(n+2)/2})$.

To show this, we know that $N(r,m)=O(logr)$ since *M* has finitely many poles.

Write (5) as

where ${A}_{1}^{\prime}=A$.

Then, there exists a constant *c* such that

Also, using (6), we can write

Also,

So,

Therefore, we get

We use Lemma 2.3 in [[13], p.38] with $R=2r$ to get

Now, we have $M=\frac{{w}^{\prime}}{w}-\frac{{v}^{\prime}}{v}$. So

We also have $N(r,M)=O(logr)$.

Hence,

This completes the proof of Claim 3.1.

Using Claim 3.1 and (14), we get

and

Also, by Theorem 1.2, we get

Therefore, we must have

and

because otherwise we can write $v=-{M}_{2}/{M}_{1}$ to get a contradiction.

We now divide (17) by *B* to get

So, ${B}^{\prime}/B$ has finitely many poles, and so *B* has finitely many zeros. Then we can write *B* in the form

where ${P}_{1}$, ${P}_{2}$ are polynomials.

But then, we can write

where ${R}_{1}$ is rational.

Then we also can write

where ${R}_{2}$ is rational and ${R}_{2}=-\frac{1}{2}{R}_{1}$.

Substitute (18) in (16), we obtain

Now, let

Also, let

So, we get

Substituting (21) in (19), we obtain

Thus, $\rho (H)<\mathrm{\infty}$, *H* is entire, and *B* has no zeros.

Then,

Therefore, *A* is a polynomial.

Since *B* has no zeros, from (20) we can write

where *Q* is a polynomial.

Since *w* and *H* solve the same equation and are linearly independent (because *w* has zeros but *H* does not), we can write

Therefore,

where ${c}_{1}$ is a constant and *Q* is a polynomial.

Now, we have

So, we can write

Therefore, we have

Hence, using (22), we obtain

where ${c}_{2}$ is a constant and ${A}_{1}^{\prime}=A$.

Now, from (23) and (24), we notice that *w* and *v* have the same zeros.

Also, differentiating (24), using (22), we have

Comparing this with (5), we get

Moreover, $H={e}^{Q}$ solves ${H}^{\u2033}+PH=0$, and so

This completes the proof of Theorem 2.1. □

## 4 A lemma needed to prove Theorem 2.2

In order to prove Theorem 2.2, we must state and prove the following lemma. We include a proof for completeness.

**Lemma 4.1** *Let* ${P}_{1},\dots ,{P}_{n}\in \mathbb{C}$ *be distinct*, *and let* ${A}_{1},\dots ,{A}_{n}$ *be rational functions such that*

*Then there exists* $k\in \{1,\dots ,n\}$ *such that* ${P}_{k}=0$ *and* ${A}_{k}=1$, *and* ${A}_{j}=0$ *for* $j\ne k$.

*Proof* The proof is by induction. It is obvious that the lemma is true when $n=1$.

Assume that the lemma is true for $m\le n-1$. Differentiating (25), we get

Now, we have two cases to consider.

Case (1): Suppose there exists *k* such that ${B}_{k}\not\equiv 0$. Without loss of generality, let $k=1$, then we can write

Since we assumed the lemma is true for $m\le n-1$, there exists $j\in \{2,\dots ,n\}$ such that ${P}_{j}-{P}_{1}=0$. But this contradicts our assumption that ${P}_{1},\dots ,{P}_{n}$ are distinct.

Case (2): Suppose that ${B}_{j}=0$ for each *j*, *i.e.*,

If ${P}_{j}\ne 0$, then ${A}_{j}\equiv 0$ because otherwise we have

But this contradicts the fact that ${P}_{j}\ne 0$.

So, we have ${A}_{j}\equiv 0$ for ${P}_{j}\ne 0$. Thus, (25) becomes (for some *k*)

and ${P}_{k}=0$ and ${A}_{k}=1$. □

## 5 Proof of Theorem 2.2

We first note that, outside a set of finite measure, by Theorem 1.1,

In particular, if *w* has finitely many zeros, then *v* is a polynomial, which gives $A=0$. This completes the proof of part (A) in the conclusion.

Assume henceforth that *w* has infinitely many zeros. Then (26) implies that $\rho (v)\le (n+2)/2$, and so *A* is a polynomial of degree at most *n* by the Wiman-Valiron theory [17]. Also, $A\not\equiv 0$ since $v-1$ has infinitely many zeros.

Now, two cases have to be considered.

Case (I). Assume that *P* is a non-zero constant; then $n=0$ and *A* is constant. Therefore, we can write

where $\alpha ,\beta \in \mathbb{C}\setminus \{0\}$, ${c}_{j},{d}_{j}\in \mathbb{C}$ and ${c}_{j}\ne 0$ ($j=1,2$).

Since *w* and $v-1$ have the same zeros with finitely many exceptions, we can write

where ${R}_{1}$ is a rational function and ${P}_{1}$ is a polynomial. We know that $deg({P}_{1})\le 1$ because $\rho (w),\rho (v)\le 1$. We can now write

where $\gamma \in \mathbb{C}$, and so

Now, by using Lemma 4.1, ${R}_{1}$ is constant and so we can write (28) as

where *δ* is constant.

Therefore,

Now, by using Lemma 4.1, we get

and $\alpha +\gamma $, $-\alpha +\gamma $, *β*, −*β*, 0 cannot all be different.

We must now try six cases:

I(a): If $\alpha +\gamma =\beta $ and $-\alpha +\gamma =-\beta $, then $\gamma =0$ and $\alpha =\beta $. But this contradicts (30). Thus, this case cannot happen.

I(b): If $\alpha +\gamma =\beta $ and $-\alpha +\gamma =0$, then $\beta =2\gamma $ and $\alpha =\gamma $. Substituting these in (30) gives

where ${h}_{1}$, ${h}_{2}$ are constants, which yields ${d}_{2}=0$. Putting this in (27) gives (7) with $\sigma =\gamma $.

There are four more cases:

I(c): $\alpha +\gamma =-\beta $ and $-\alpha +\gamma =\beta $.

I(d): $\alpha +\gamma =-\beta $ and $-\alpha +\gamma =0$.

I(e): $\alpha +\gamma =0$ and $-\alpha +\gamma =\beta $.

I(f): $\alpha +\gamma =0$ and $-\alpha +\gamma =-\beta $.

It is easy to check that case I(c) is impossible and cases I(d), I(e), I(f) all lead to (7) with $\sigma =\gamma $.

From these cases, we find that $\gamma \ne 0$, and so $\frac{v-1}{w}={e}^{\gamma z+\delta}$ is non-constant. Also, we have (7) and case (B) of the conclusion.

Case (II). Suppose that *P* is non-constant. We will show that this leads to a contradiction. Let

where *L* is a rational function and *Q* is an entire function.

From (26), we have $\rho (v)<\mathrm{\infty}$, and so *Q* is a polynomial.

Also, from (31), we have

So,

Now, we have two cases to consider.

Case (i): If *M* is constant, then either $A=P$ and $A=0$, so that $P=0$, which is a contradiction, or

is a rational function, which is a contradiction since *w* has infinitely many zeros.

Case (ii): If *M* is non-constant, then ${M}^{\prime}\not\equiv 0$. Therefore,

where $\frac{{M}^{\u2033}}{2{M}^{\prime}}+\frac{(A-P)M}{2{M}^{\prime}}$ is rational because $\frac{{M}^{\prime}}{M}=\frac{{L}^{\prime}}{L}+{Q}^{\prime}$ is rational and $\frac{{M}^{\u2033}}{M}$ is rational, and so $\frac{{M}^{\u2033}}{{M}^{\prime}}=\frac{{M}^{\u2033}}{M}/\frac{{M}^{\prime}}{M}$ is rational.

Also,

Then we can write (33) as

where

are rational functions and *Q* is a polynomial.

Let $U=S{e}^{-Q}$, then we can write (34) as

Now, we have two cases to consider:

Case ii(a): If $R\equiv 0$ in (35), then (34) gives

which is a contradiction since *w* has infinitely many zeros.

Case ii(b): Assume that $R\not\equiv 0$ in (35); then (35) gives

Now, (1) and (35) give

and so

Therefore,

because if not, *w* has finitely many zeros, a contradiction. Also,

Put

where $T=1/S$ is a rational function.

Then,

From (36), we get

So, *G* solves (1), and since $P\not\equiv 0$ and is a polynomial of degree *n*, we see that *G* is a transcendental entire function with finitely many zeros and has order $(n+2)/2$.

Since *w* and *G* solve the same equation but *w* has infinitely many zeros and *G* has finitely many zeros, *w* and *G* are linearly independent, and we can write

where *c* is a non-zero constant. So,

By integrating, we get

Also, using (31), (38) and (39),

where $H=L/T$ is a rational function.

Now, we can assume that $c=1$ because if $c\ne 1$, we can multiply *w* by $1/c$.

We differentiate (40) to get

where

is a rational function.

So, from (3) and (41), we get

and so

where ${U}_{1}={K}^{\prime}+{K}^{2}+A$ and ${U}_{2}={H}^{\prime}-{K}^{\prime}+KH-{K}^{2}$.

Since *v* is transcendental and ${U}_{1}$, ${U}_{2}$ are rational functions, we must have

and

**Claim 5.1** *We claim that* $H\equiv K$.

To show this, let $H\not\equiv K$.

From (44), we have

From (42), we get

We integrate to get

where *a* is a constant. But this contradicts the fact that *H* and *K* are rational functions and *G* is a transcendental function. This completes the proof of Claim 5.1.

Once we have Claim 5.1, (41) gives

and so

By (45), *v* has finitely many zeros, so we can write

where ${P}_{1}$, ${Q}_{1}$ are polynomials, ${P}_{1}\not\equiv 0$, and ${Q}_{1}$ is non-constant because *v* is transcendental.

Therefore,

Now, we can write this as

where ${R}_{1}={P}_{1}/L\not\equiv 0$, ${R}_{2}=-1/L\not\equiv 0$ are rational functions and ${S}_{1}={Q}_{1}-Q$, ${S}_{2}=-Q$ are polynomials.

Here, ${R}_{1}{e}^{{S}_{1}}$ and ${R}_{2}{e}^{{S}_{2}}$ are linearly independent because ${Q}_{1}$ is non-constant. Now, we get

where ${J}_{1}$, ${J}_{2}$ are rational and satisfy

Therefore, ${J}_{1}={J}_{2}=0$ because otherwise ${e}^{{Q}_{1}}=-{J}_{2}/{J}_{1}$ or ${e}^{-{Q}_{1}}={J}_{1}/{J}_{2}$. Thus, ${R}_{1}{e}^{{S}_{1}}$, ${R}_{2}{e}^{{S}_{2}}$ both solve ${y}^{\u2033}+Py=0$ and have finitely many zeros, and they are linearly independent.

Hence, *P* is constant by [9], which contradicts our assumption in Case (II) that *P* is non-constant. □

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## Acknowledgements

The author would like to thank his supervisor Prof. Jim Langley for his support and guidance. Also, he would like to thank King Abdulaziz University for financial support for his PhD study.

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Asiri, A. Further results on common zeros of the solutions of two differential equations.
*J Inequal Appl* **2012, **222 (2012). https://doi.org/10.1186/1029-242X-2012-222

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### Keywords

- Nevanlinna theory
- differential equations